June 2013 - mathfun@mailman.xmission.com

[math-fun] Chunks to be erased = summing digits
by Eric Angelini 21 Jun '13

21 Jun '13

Hello Math-Fun, The idea is to read the decimalization of Pi from left to right and to delete _all_ the read digits with the hereunder rule: - Sum, one by one, the digits you encounter; - When the partial sum is visible immediately to the right of the summed digits, erase the said digits; - Start summing again, from zero. Example: Pi decimalization starts: 3.1.4.1.5.9.2.6.5.3.5.8.9.7.9.3.2.3.8.4.6.2.6.4.3.3.8.3.2.7.9.5.0.2.8.8.4.1.9.7.1.6.9.3.9.9.3.7.5.1.0.5.8.2.0.9.7.4.9.4.4.5.9.2.3.0.7.8.1.6.4.0.6.2.8.6.2.0.8.9.9.8.6.2.8.0.3.4.8.2.5.3.4.2.1.1.7.0.6.7.9... We read from left to right; as 3+1 = 4 we erase the first two digits; we are left with: 4.1.5.9.2.6.5.3.5.8.9.7.9.3.2.3.8.4.6.2.6.4.3.3.8.3.2.7.9.5.0.2.8.8.4.1.9.7.1.6.9.3.9.9.3.7.5.1.0.5.8.2.0.9.7.4.9.4.4.5.9.2.3.0.7.8.1.6.4.0.6.2.8.6.2.0.8.9.9.8.6.2.8.0.3.4.8.2.5.3.4.2.1.1.7.0.6.7.9... We start summing again, from left to right; as 4+1 = 5 we erase the first two digits; we are left with: 5.9.2.6.5.3.5.8.9.7.9.3.2.3.8.4.6.2.6.4.3.3.8.3.2.7.9.5.0.2.8.8.4.1.9.7.1.6.9.3.9.9.3.7.5.1.0.5.8.2.0.9.7.4.9.4.4.5.9.2.3.0.7.8.1.6.4.0.6.2.8.6.2.0.8.9.9.8.6.2.8.0.3.4.8.2.5.3.4.2.1.1.7.0.6.7.9... We start summing again, from left to right; as ... [and here I must confess that I don't have the tools to find the next chunk of digits; the sum is bigger than 100] This erasing technique can be summarized by the size (in digits) of each erased chunk. We would thus have(for Pi): [2,2,...] Best, É. e size of each chunk is the signature of th

2 3

[math-fun] book on Homotopy Type Theory
by Ray Tayek 20 Jun '13

20 Jun '13

http://audrey.fmf.uni-lj.si/hott.html NOTE: my blog is being slashdotted by Hacker News. Please read this static version of the blog post while I resolve the issue. By Andrej Bauer, on June 20th, 2013 The HoTT book is finished! Since spring, and even before that, I have participated in a great collaborative effort on writing a book on Homotopy Type Theory. It is finally finished and ready for public consumption. You can get the book freely at <http://homotopytypetheory.org/book/>http://homotopytypetheory.org/book/. Mike Shulman has written <http://golem.ph.utexas.edu/category/2013/06/the_hott_book.html>about the contents of the book, so I am not going to repeat that here. Instead, I would like to comment on the socio-technological aspects of making the book, and in particular about what we learned from open-source community about collaborative research. We are a group of two dozen mathematicians who wrote a 600 page book in less than half a year. This is quite amazing, since mathematicians do not normally work together in large groups. In a small group they can get away with using obsolete technology, such as sending each other source LaTeX files by email, but with two dozen people even Dropbox or any other file synchronization system would have failed miserably. Luckily, many of us are computer scientists disguised as mathematicians, so we knew how to tackle the logistics. We used <http://git-scm.com/>git and <https://github.com/>github.com. In the beginning it took some convincing and getting used to, although it was not too bad. In the end the repository served not only as an archive for our files, but also as a central hub for planning and discussions. For several months I checked github more often than email and Facbook. Github was my Facebook (without the cute kittens). If you do not know about tools like git but you write scientific papers (or you create any kind of digital content) you really, really should learn about <https://en.wikipedia.org/wiki/Revision_control>revision control systems. Even as a sole author of a paper you will profit from learning how to use one, not to mention that you can make <https://vimeo.com/68761218>pretty videos of how you wrote your paper. ... --- co-chair http://ocjug.org/

1 0

[math-fun] Fwd: Off topic delight
by Whitfield Diffie 14 Jun '13

14 Jun '13

Yesterday, John Young sent me the following pointer http://techpinions.com/an-old-mystery-solved-project-c-43-and-public-key-en… I think we have been looking for this for fifteen years. Whit

1 0

[math-fun] Zhang's proof of the bounded prime gap theorem
by Victor Miller 13 Jun '13

13 Jun '13

Emanuel Kowalski's blog gives a useful summary: http://blogs.ethz.ch/kowalski/2013/05/21/bounded-gaps-between-primes/

6 8

[math-fun] Maple 1, Mma 1
by Fred lunnon 12 Jun '13

12 Jun '13

Via http://www.wolframalpha.com , an admittedly unreliable source ... In: (Binomial[n-1, k-1] + Binomial[n-1, k] - Binomial[n, k]) Out: 0 In: (Binomial[n-1, k-1] + Binomial[n-1, k] - Binomial[n, k]) /. {n -> 0, k -> 0} Out: 1 Oops! Maple, on the other hand, discreetly if uninformatively desists from further simplification of the general expression. But not merely from cowardice: given the Catalan number recurrence, which Wolfa either claims implausibly not to understand, or else goes all coy over and refuses to reply at all: In: (Binomial[n+n, n]/(n+1) - Sum[(Binomial[j+j-2, j-1]/j)(Binomial[n+n-j-j, n-j]/(n-j+1)), {j, 1, n}]) Out: <caposcalpitation / mysterious silence> Maple in contrast fearlessly extends its neck: In: n := 'n'; simplify( binomial(n+n, n)/(n+1) - sum(binomial(j+j-2, j-1)/j*binomial(n+n-j-j, n-j)/(n-j+1), j = 1..n) ); Out: 0 In: n := 0; binomial(n+n, n)/(n+1) - add(binomial(j+j-2, j-1)/j*binomial(n+n-j-j, n-j)/(n-j+1), j = 1..n); Out: 1 and duly measures its length on the carpet (which one way and another seems to be having a hard time of it lately). Would you buy a used hypergeometric identity from any of these head-CASes? Fred Lunnon

1 0

[math-fun] Multiple proofreaders
by Bernie Cosell 10 Jun '13

10 Jun '13

In a [very] old issue of the MAA journal, there was a short article about estimating the number of errors still remaining after something had been checked by two proofreaders. Does anyone recall that? /Bernie\ -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--

3 3

[math-fun] Lore of the Rings
by Hans Havermann 09 Jun '13

09 Jun '13

I've started a very non-mathematical, historical investigation of Pacioli's Rings puzzle (if I may call it that). My main purpose (apart from the joy of exploration) was to collect in one place a lot of source material, imbue the subject matter with some general sense of skepticism, and posit a few queries: http://gladhoboexpress.blogspot.ca/2013/06/lore-of-complicated-rings.html

1 0

[math-fun] Bracketings, ORTs and binomial coefficients
by Fred lunnon 08 Jun '13

08 Jun '13

Now why did I want to generate n-bracketings in the first place ... or what amounts to the same thing, ordered rooted trees (ORTs) with n+1 nodes? [Here each node corresponds to a matching pair of brackets, and its children to the pairs immediately within the parent pair. An ORT will be equipped with an extra half-edge ("bole"?) leading into the root node, which corresponds to a virtual outer pair of brackets imagined enclosing any given bracketing.] How many k-valent nodes occur among the set of all distinct (n+1)-ORTs? After spending all that time generating the damn' things, the answer is an anticlimax: apparently it's just (2n-k)_C_(n-1) the dear old binomial coefficient! Is this well-known? Can anyone suggest a proof? If so, should OEIS A007318 be amended to incorporate: " n_C_k = number of (2k+2-n)-valent nodes occurring among distinct ordered rooted trees with k+2 nodes, provided each root node is equipped with an extra inward half-edge " ? Example: n = 3 bracketing valency 1,...,4 ( ((())) ) [1, 3, 0, 0] ( (()()) ) [2, 1, 1, 0] ( ()(()) ) [2, 1, 1, 0] ( (())() ) [2, 1, 1, 0] ( ()()() ) [3, 0, 0, 1] totals [10, 6, 3, 1] Fred Lunnon

1 1

Re: [math-fun] squared squares and squared rectangles (Stuart Anderson)
by Stuart Anderson 08 Jun '13

08 Jun '13

I had a look at James William's simple perfect squared squares (SPSSs) collection (15 million) and found some new record holder in higher orders. In order 41 and above several record breaking 'pairs' showed up. There is a pair in order 43 with a side of 35172, and largest/smallest elements of 10558 and 1312; 10558/1312=8.05. A pdf with 30 of the best is here; http://www.squaring.net/downloads/nice-spss-JBW.pdf I also had a look in simple perfect squared rectangles (SPSRs) all the way up to order 24. There are 63,301,415 in order 23 and 228,130,900 in order 24. In order 23 I found a SPSR 7526x5620 with largest/smallest elements of 2182 and 576; 2182/576 = 3.788. This was the only one with a ratio below 4, but there were many with 4 and above. A pdf is here; http://www.squaring.net/downloads/23-7526x5620-3.788.pdf So it looks like the ratio will approach 1 for SPSRs and I think this is probably the case for SPSSs too, though much higher orders would be required. A related problem is to find the maximum possible ratio of the LARGEST element in a squared square to the side of the squared square. In 2009 Brian Trial found 1xn SPSRs up to 1x18. The 1xn squared rectangle is a generalisation of squared squares and a harder problem, the squared domino (1x2) didnt appear until order 22, while the lowest order SPSS found by Duijvestijn was order 21 (112 side). Up until 2009 only one example of a 1x3 SPSR was known, due to Jasper Skinner, found in order 26, so Brian's finds were quite extraordinary. Two pdfs of Brian's 1xn SPSRs are here; (landscape version) http://www.squaring.net/sq/sr/B_T-18b.pdf and (portrait version) http://www.squaring.net/sq/sr/B_T-18a.pdf By scaling and fitting Brian's 1x17 and 1x18 SPSRs and adding one extra square, equal to the size of the long side of the scaled 1x17 rectangle, a compound perfect squared square (CPSS) is created with the largest element square having a side equal to 94% of the CPSS side. So it seems the maximum of largest element to squared square size also approaches 1, from below this time. A pdf of this CPSS is here http://www.squaring.net/downloads/560-338321736_BT.pdf , you need to zoom in to see the elements along the sides. Stuart Anderson > > Message: 3 > Date: Tue, 4 Jun 2013 20:30:11 -0700 > From: Bill Gosper <billgosper(a)gmail.com> > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] squared squares > Message-ID: > <CAA-4O0FEeaXO0B3PRJ2Mi28Em5V7LzGTJ1my7MNo78gjgpwfTQ(a)mail.gmail.com> > Content-Type: text/plain; charset=windows-1252 > > Stuart Anderson>Armin Singer has written a program to search for > 'nice' squared squares > http://www.squaring.net/downloads/squaredance-0.10.tgz > He has a couple of 'nice' files in the tarball > Also James Williams produced over 15 million new perfect squared > squares earlier this year. Most of them havnt been added to the > website yet. There are bound to be a whole lot of record breakers in > that lot. > There are some 'nice' squared rectangles here > http://www.squaring.net/sq/sr/spsr/spsr_minmax.html > > Stuart > > While I don't see much nice in nice2.txt, the new record in nice1 > appears to be 65/8: > > ShowQuilt[{31, 1408, 1408, 448, 520, 440, 80, 360, 262, 186, 114, 239, > 247, 175, 125, 163, 99, 260, 104, 96, 151, 118, 242, 64, 210, 200, > 227, 145, 124, 81, 389, 366, 345, 308}, True] > > where, from Ed Pegg's MondrianPuzzles.nb > (*Code for generating positions of squares based on a Bouwkamp code.*) > > BouwkampCoords[BouwkampFormat_List] := > Module[{nums = BouwkampFormat}, > {order, height, width } = Take[nums, 3]; > squares = Drop[nums, 3]; > active = {{{0, 0}, {0, height}}} ; > coordinates = Table[{0}, {order}]; > cornera = squares[[1]]; > cornerb = cornerc = cornerd = 0; > Do[ > ss = squares[[n]]; > coordinates[[n]] = > Table[active[[1, 1]], {4}] + {{0, 0}, {0, ss}, {ss, ss}, {ss, > 0}}; > active = > Sort[Switch[Sign[active[[1, 1, 2]] + ss - active[[1, 2, 2]]], > 0, ReplacePart[active, 1 -> active[[1]] + {{ss, 0}, {ss, 0}}], > -1, > Flatten[{Drop[active, > 1], {active[[1]] + {{0, ss}, {0, 0}}, {active[[1, 1]], > active[[1, 1]]} + {{ss, 0}, {ss, ss}} }}, 1], > _, Flatten[{active, {{"Error", n}}}, 1]]]; > If[active[[1, 1]] == "Error", Break[]]; > active = > Partition[ > Flatten[Select[Split[Flatten[active, 1]], Length[#] < 2 &], 1], > 2]; > If[coordinates[[n, 2]] == {0, height}, cornerb = squares[[n]]]; > If[coordinates[[n, 3]] == {width, height}, > cornerc = squares[[n]]]; > If[coordinates[[n, 4]] == {width, 0}, cornerd = squares[[n]]], > {n, 1, order}]; > coordinates = If[active == {{{width, 0}, {width, height}}}, > Append[ > First /@ coordinates, {cornera, cornerb, cornerc, cornerd}], > Append[ > First /@ coordinates, {cornera, cornerb, cornerc, cornerd, > active}]]] > > (*Display squares of a Bouwkamp code.*) > > ShowQuilt[BouwkampFormat_List, color_: False] := > With[{ > vals = Drop[BouwkampFormat, 3], order = BouwkampFormat[[1]], > size = BouwkampFormat[[2]], > coord = BouwkampCoords[BouwkampFormat]}, > With[{colors = > If[color, {Gray, White, Red, Yellow, Blue, Cyan, Magenta}[[ > Mod[#, 7, 1]]] & /@ vals, Table[White, {Length[vals]}]]}, > Graphics[{EdgeForm[{Black, Thick}], Opacity[.3], > Table[{colors[[n]], > Rectangle[coord[[n]], coord[[n]] + vals[[n]]]}, {n, 1, order}], > Opacity[1], Black, > Table[ > If[vals[[n]] > size/15, > Style[Text[vals[[n]], coord[[n]] + vals[[n]]/2], 18], > Sequence @@ {}], {n, 1, order}]}, ImageSize -> {450, 370}]]] > > I wonder if the min over all squared squares exists and is > 1. --rwg > > > On Tue, Jun 4, 2013 at 9:57 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > Ed> In the order 29 squares, a square of size 1702 has Smallest square = >> 62Largest Square = 567 >> That's a ratio of 9.1 >> --Ed Pegg Jr >> Wow. >> >> --- On Tue, 6/4/13, Dan Asimov <dasimov(a)earthlink.net> wrote: >> >> From: Dan Asimov <dasimov(a)earthlink.net> >> Subject: Re: [math-fun] squared squares >> To: "math-fun" <math-fun(a)mailman.xmission.com> >> Date: Tuesday, June 4, 2013, 8:44 AM >> >> I see that Ian Gambini describes a squared square of side 110 in >> >> Gambini, Ian A method for cutting squares into distinct squares. >> Discrete Appl. >> Math. 98 (1999), no. 1-2, 65?80, >> >> with a total of 23 squares whose sides range from s=2 to s=44, for a >> ratio of 22. >> >> --Dan >> No, there's a 1x1 in there. But a later example has 50/2. >> > I was just looking at a story where one Rodolfo Gambini uses quantum > gravity to exonerate God from dividing by zero. Small legs must confer > a selective advantage. > >> (There's an image of the squared square here: >> < http://www.sciencedirect.com/science/article/pii/S0166218X99001584 >, >> but it may require a subscription.) >> >> >> On 2013-06-03, at 11:11 PM, Bill Gosper wrote: >> >> http://www.squaring.net/sq/ss/spss/o22/spsso22.pdf has >> largest/smallest = 30. Is this minimal? What about squared >> rectangles? --rwg _______________________________________________ >> Tnx! >> --rwg >

2 1

Re: [math-fun] Naturally-occurring banana shaped like an integral sign
by Adam P. Goucher 08 Jun '13

08 Jun '13

I was referring to the curvature of the curve running axially through the banana, rather than the surface of the banana. Sincerely, Adam P. Goucher > ----- Original Message ----- > From: Phil Carmody > Sent: 06/08/13 09:27 AM > To: math-fun(a)mailman.xmission.com > Subject: Re: [math-fun] Naturally-occurring banana shaped like an integral sign > > From: "Adam P. Goucher" <apgoucher(a)gmx.com> > > The title is pretty self-explanatory. > > > > http://cp4space.wordpress.com/2013/06/05/integral-banana/ > > "It is renowned for having a constant positive curvature" ?!??! > > If you've not noticed the saddle, you've clearly never ridden on a giant inflatable one! > > Phil > > -- > () ASCII ribbon campaign () Hopeless ribbon campaign > /\ against HTML mail /\ against gratuitous bloodshed > > [stolen with permission from Daniel B. Cristofani] > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

1 0